Prove value of $\sin(15°)$

They are equivalent. Since $$(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 2(2 + \sqrt{3}),$$ it follows that $$\sqrt{2 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{2}}.$$ Then $$\frac{1}{2 \sqrt{2 + \sqrt{3}}} = \frac{\sqrt{2}}{2 (1 + \sqrt{3})} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}.$$

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It is also worth noting that we can obtain the desired expression more directly via the half-angle identity for $0 \le \theta \le 90^\circ$: $$\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \frac{\sqrt{1 - \cos^2 \theta}}{\sqrt{2 (1 + \cos \theta)}} = \frac{\sin \theta}{\sqrt{2 (1 + \cos \theta)}},$$ where upon selecting $\theta = 30^\circ$ immediately yields $$\sin 15^\circ = \frac{1}{2 \sqrt{2 + \sqrt{3}}}$$

Since, ${\sqrt3 - 1} = \sqrt{(\sqrt3 - 1)^2}= \sqrt2 \sqrt{2-\sqrt3}$ $$\frac{\sqrt3 - 1}{2\sqrt2} = \frac{\sqrt{2 - \sqrt3}}{2}$$ Rationalising the numerator $$\frac{\sqrt{2 - \sqrt3}}{2}\cdot \frac{\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}= \frac{1}{2\sqrt{2+\sqrt3}}$$